Question

The volume of a sphere is increasing at a rate of 6π cubic centimeters per hour. At what rate, in centimeters per hour, is its diameter increasing with respect to time at the instant the radius of the sphere is 3 centimeters.

A: 1/3
B: 1
C: √6
D: 6

Answer 1

Answer:  Diameter is increasing as $\frac{1}{3}$ centimeter per hour.

Step-by-step explanation:

Alright, lets get started.

The formula for volume of sphere is given as V : $\frac{4}{3}\pir^3$

The volume of a sphere is increasing at a rate of $6\pi$ cubic centimeters per hour.

It means : $\frac{dV}{dt}=6\pi$

$V=\frac{4}{3}\pi r^3$

Taking derivative with respect to t

$\frac{dV}{dt}=\frac{4}{3}\pi \times 3r^2 \frac{dr}{dt}$

$6\pi=4\pi r^2 \frac{dr}{dt}$

at the instant the radius of the sphere is 3 centimeters, means

$\frac{dr}{dt}= \frac{6}{4 \times 3^2}$

$\frac{dr}{dt}=\frac{1}{6}$

As $radius = \frac{diameter}{2}$

$\frac{1}{2}\frac{dD}{dt}=\frac{1}{6}$

$\frac{dD}{dt} =\frac{1}{3}$

Hence diameter is increasing as $\frac{1}{3}$ centimeter per hour.   : Answer

Hope it will help :)

Answer 2

The volume of a sphere is increasing at a rate of 6π cubic centimeters per hour. The rate at which its diameter is increasing with respect to time at the instant with the radius of the sphere 3 centimeters is: $\mathbf{\dfrac{1}{3}}$

Option A is correct.

The volume of a sphere can be represented by using the formula:

$\mathbf{V = \dfrac{4}{3}\pi r^3}$

Now, by differentiation, if we differentiate the rate at which the volume is increasing with time, we have:

$\mathbf{\dfrac{dV}{dt} = \dfrac{4}{3}\pi r^3 \ \dfrac{dr}{dt}}$

$\mathbf{\dfrac{dV}{dt} = 4 \pi r^2 \ \dfrac{dr}{dt}}$

Given that:

• $\mathbf{\dfrac{dV}{dt}= 6 \pi cm^3/ sec}$
• radius (r) = 3 cm

Replacing the values into the differentiated equation, we have:

$\mathbf{6 \pi= 4 \pi (3)^2 \ \dfrac{dr}{dt}}$

$\mathbf{\ \dfrac{dr}{dt}=\dfrac{6 \pi}{4 \pi (3)^2}}$

$\mathbf{\ \dfrac{dr}{dt}=\dfrac{6 }{4 \times 9}}$

$\mathbf{\ \dfrac{dr}{dt}=\dfrac{1 }{6}\ cm/sec}$

• Recall that radius = d/2

∴

$\mathbf{\dfrac{1}{2} \dfrac{dr}{dt} =\dfrac{1}{6} }$

$\mathbf{ \dfrac{dr}{dt} =\dfrac{1}{6} \times \dfrac{2}{1}}$

$\mathbf{ \dfrac{dr}{dt} =\dfrac{1}{3}}$

Therefore, we can conclude that the rate at which its diameter is increasing with respect to time at the instant with the radius of the sphere 3 centimeters is  $\mathbf{\dfrac{1}{3}}$

Learn more about spheres here:

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